Lebesgue differentiation theorem on ${\mathbb{R}}^n$

For each integer $m\ge 1$ and each integer $j\ge 1$, apply [L2] to the [L2, given, choose, construct] $L^1$ function $f{\mathbf{1}}_{B(0,m+1)}$ and choose $g_{m,j}\in C_c({\mathbb{R}}^n)$ such that ${\int }_{B(0,m+1)}\mathrm{\mid }f-g_{m,j}\mathrm{\mid }d\lambda <\frac{2^{-2j}}{5^n}\mathrm{.}$ Set $h_{m,j}:=(f-g_{m,j}){\mathbf{1}}_{B(0,m+1)}\mathrm{.}$

Let [L3, L4, L5, step 1.1, algebra] $E_{m,j}:=\{x\in B(0,m):M{h}_{m,j}(x)>2^{-j}\}\cup \{x\in B(0,m):\mathrm{\mid }{h}_{m,j}(x)\mathrm{\mid }>2^{-j}\}\mathrm{.}$ Both sets are Lebesgue measurable: the maximal superlevel set is open by the ball-average continuity argument in [L4], and $h_{m,j}$ is measurable. Finite subadditivity [L5], [L4] and [L3] give $\lambda (E_{m,j})\le \frac{5^n}{2^{-j}}\mathrm{\parallel }{h}_{m,j}{\mathrm{\parallel }}_1+\frac{1}{2^{-j}}\mathrm{\parallel }{h}_{m,j}{\mathrm{\parallel }}_1<2^{-j}+5^{-n}{2}^{-j}\le 2^{1-j}\mathrm{.}$

For fixed $m$, put [step 2.1, algebra] $N_m:={\bigcap }_{N=1}^{\mathrm{\infty }}{\bigcup }_{j\ge N}{E}_{m,j}\mathrm{.}$ The sets $E_{m,j}$, their countable unions, and $N_m$ are measurable by [L5]. By countable subadditivity [L5] and step 2.1 $\lambda \text{\hspace{0.17em}⁣}\left({\bigcup }_{j\ge N}{E}_{m,j}\right)\le {\sum }_{j\ge N}\lambda (E_{m,j})\le {\sum }_{j\ge N}{2}^{1-j}=2^{2-N}\mathrm{.}$ For every $N$, inclusion and monotonicity [L5] give $0\le \lambda (N_m)\le 2^{2-N}$. Given $ϵ>0$, choose $N$ with $2^{2-N}<ϵ$; thus $\lambda (N_m)=0$. This uses small positive tail measures, without asserting that any $E_{m,j}$ is null.

Let $x\in B(0,m)\setminus N_m$. Then there is $N_x$ such that [L1, step 1.1, step 3.1, algebra] $x\mathrm{\notin }{E}_{m,j}$ for every $j\ge N_x$. Fix such a $j$ and take $0<r<1$. Since $x\in B(0,m)$, one has $B(x,r)\subseteq B(0,m+1)$, so $A_{r}f(x)-f(x)=(A_r{g}_{m,j}(x)-g_{m,j}(x))+A_r{h}_{m,j}(x)-h_{m,j}(x)\mathrm{.}$ Therefore $\mathrm{\mid }{A}_{r}f(x)-f(x)\mathrm{\mid }\le \mathrm{\mid }{A}_r{g}_{m,j}(x)-g_{m,j}(x)\mathrm{\mid }+M{h}_{m,j}(x)+\mathrm{\mid }{h}_{m,j}(x)\mathrm{\mid }\le \mathrm{\mid }{A}_r{g}_{m,j}(x)-g_{m,j}(x)\mathrm{\mid }+2^{1-j}\mathrm{.}$ Now [L1] gives $A_r{g}_{m,j}(x)\to g_{m,j}(x)$, so ${\mathrm{lim\hspace{0.17em}sup}}_{r\to 0^{+}}\mathrm{\mid }{A}_{r}f(x)-f(x)\mathrm{\mid }\le 2^{1-j}$ for every $j\ge N_x$. Letting $j\to \mathrm{\infty }$ yields $A_{r}f(x)\to f(x)$.

The bad set for radius differentiation is contained in [step 3.1, step 4.1, L5] ${\bigcup }_{m\ge 1}{N}_m$, which is null by [L5]. Thus the convergence holds for almost every $x\in {\mathbb{R}}^n$.